To find a precise expression for the sum of the squares of the first [tex]\( n \)[/tex] natural numbers [tex]\( 1^2 + 2^2 + 3^2 + \cdots + n^2 \)[/tex], we follow these steps:
1. Define the Sum: Let [tex]\( S \)[/tex] represent the sum of the squares of the first [tex]\( n \)[/tex] natural numbers.
[tex]\[
S = 1^2 + 2^2 + 3^2 + \cdots + n^2
\][/tex]
2. Objective:
We need to simplify [tex]\( S \)[/tex] and find the formula for it.
3. Known Formula for Sum of Squares:
The formula for the sum of the squares of the first [tex]\( n \)[/tex] natural numbers is:
[tex]\[
S = \frac{n (n + 1) (2n + 1)}{6}
\][/tex]
4. Proof of the Formula:
To verify the formula, we can calculate a closed form for the summation:
[tex]\[
S = \sum_{k=1}^n k^2
\][/tex]
5. Simplification Step:
Using mathematical tools (such as induction or polynomial fitting, which are methods you'd learn in more advanced courses), we know that this sum simplifies to:
[tex]\[
S = \frac{n (n + 1) (2n + 1)}{6}
\][/tex]
6. Comparison Expression:
We also know another form from polynomial expansion:
Expanding [tex]\(\frac{n (n + 1) (2n + 1)}{6} \)[/tex], it matches with:
[tex]\[
\frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}
\][/tex]
Thus, equating our sum and the expanded polynomial form, we find:
[tex]\[
S = \frac{n (n + 1) (2n + 1)}{6} = \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}
\][/tex]
Therefore, the sum of the squares of the first [tex]\( n \)[/tex] natural numbers can indeed be represented by the formula:
[tex]\[
1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n (n + 1) (2n + 1)}{6}
\][/tex]
So, when asked whether [tex]\( 1^2 + 2^2 + 3^2 + \cdots + n^2 = n(n+1)(2n+1) \)[/tex], it would be incorrect to say they are equal directly without the division by 6. Hence, the correct formula is:
[tex]\[
1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n (n + 1) (2n + 1)}{6}
\][/tex]
